I have a question about Brownian motion.
Let $(\Omega,\mathcal{F},P)$ be a Probability space and $(B_{t})_{t \in [0,\infty[}$ be a standard $1$-dimensional Brownian motion defined on $(\Omega,\mathcal{F},P)$.
Fix $x \in \mathbb{R}$. Brownian motion started at $x$ induces a Probability measure $P_{x}$ on the path space $(W,\mathcal{B}(W)):=(C([0,\infty[),\mathcal{B}(C([0,\infty[)))$ by $P_{x}(A)=P(\{\omega \in \Omega : B_{\cdot}(\omega)+x \in A \})\quad(A \in \mathcal{B}(C([0,\infty[)))$.
My question:
For any bounded continuous function $F:W\to \mathbb{R}$, the function $\mathbb{R} \ni x \to E_{x}[F]:=\int_{W}F(w)P_{x}(dw) $ is continuous (not measurable)?
My attempt:
For $F(w)=\prod_{k=1}^{n}f_{k}(\pi_{t_{k}}(w))$ (where $f_{k}:\mathbb{R}\to \mathbb{R}$, bounded countinous, $\pi_{t_{k}}:W \ni w \mapsto w(t_{k}) \in \mathbb{R}$, $0 \leq t_{1} \leq t_{2} \leq \ldots \leq t_{n} < \infty$) \begin{align*} E_{x}\left[F \right]=E \left[\prod_{k=1}^{n}f_{k}(B_{t_{k}}+x)\right]\quad(x \in \mathbb{R}) \end{align*} Since each $f_{k}$ is bounded continuous function, $x \mapsto E_{x}[F]$ is continuous.
Thank you in advance.
It holds that
$$\mathbb{E}^x(F) = \int F(w) \, d\mathbb{P}_x(w) = \int F(x+w) \, d\mathbb{P}(w) \tag{1}$$
for any measurable function $F: (C[0,\infty),\mathcal{B}(C[0,\infty)) \to [0,\infty)$. This follows from the fact that $(1)$ holds for simple functions and we can extend the equality to all measurable non-negative functions (by the monotone convergence theorem). Finally, we can extend $(1)$ to all measurable functions $F$ such that the right-hand side is well-defined (e.g. if $F$ is bounded).
Applying $(1)$ to the bounded continuous (hence measurable) function $F$ yields
$$\mathbb{E}^x(F) = \int F(x+w) \, d\mathbb{P}.$$
Since $x \mapsto F(x+w)$ is continuous, it follows directly from the dominated convergence theorem that $x \mapsto \mathbb{E}^x(F)$ is continuous.